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anonymous

  • one year ago

The claim is that the proportion of peas with yellow pods is equal to 0.25 (or 25%). The sample statistics from one experiment include 570 peas with 111 of them having yellow pods. Find the value of the test statistic.

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  1. anonymous
    • one year ago
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    hi

  2. anonymous
    • one year ago
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    Hey there

  3. anonymous
    • one year ago
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    hi

  4. anonymous
    • one year ago
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    don`t understand the question!!!

  5. anonymous
    • one year ago
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    Nope

  6. amistre64
    • one year ago
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    what do you not understand about the question?

  7. anonymous
    • one year ago
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    How to put the equation into my TI83 to obtain the value

  8. anonymous
    • one year ago
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    Of the test statistic

  9. amistre64
    • one year ago
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    looks like we have 1 sample, and are tring to evaluate a proportion ...

  10. amistre64
    • one year ago
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    what does 2nd, vars, tests (maybe calc) get for us?

  11. amistre64
    • one year ago
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    http://cfcc.edu/faculty/cmoore/TI-STAT.htm

  12. anonymous
    • one year ago
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    ok I have it up but how would I input the information?

  13. amistre64
    • one year ago
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  14. amistre64
    • one year ago
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    tell me what the inputs are ... are what youd assume they should be

  15. amistre64
    • one year ago
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    po = ? n = ? x = ?

  16. anonymous
    • one year ago
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    po=0.25 n=600 x= no sure about

  17. anonymous
    • one year ago
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    not 600 I meant 570

  18. amistre64
    • one year ago
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    po is good, now from the information we have only 2 numbers left to work 570 peas with 111 of them having yellow pods

  19. anonymous
    • one year ago
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    Yes. So, then to put that into an equation is what I am having difficulty with.

  20. amistre64
    • one year ago
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    there is no equation, you simply type them into the calculator po = .25 n=570 x=111 the last thing to do is to determine the type of tails we want to test.

  21. amistre64
    • one year ago
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    you can work it as an equation, but then why would we have used the function to start with?

  22. anonymous
    • one year ago
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    Oh ok... So how would I input it into the ti83?

  23. amistre64
    • one year ago
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    i already covered that ... with links and pictures and everything.

  24. anonymous
    • one year ago
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    Ok, I will try it out.

  25. amistre64
    • one year ago
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    once you get to the function and fill in those 3 parts, theres one last part to cover. let me know when you are there.

  26. anonymous
    • one year ago
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    ok, I have it calculated

  27. amistre64
    • one year ago
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    and what is your value?

  28. amistre64
    • one year ago
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    also, what did you choose for the last portion? \(\ne\)p, <p, or >p

  29. anonymous
    • one year ago
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    All I see is: z=-3.047 p=.0011 p^=.1947 n=570

  30. amistre64
    • one year ago
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    your test stat is z :)

  31. anonymous
    • one year ago
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    I just clicked on calculate

  32. amistre64
    • one year ago
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    thats fine, we werent asked anything that would alter the z value.

  33. anonymous
    • one year ago
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    OMG thank you thank you!!!

  34. amistre64
    • one year ago
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    another way to have done is: z = (111/570 - .25)/sqrt(.25(1-.25)/570) which is -3.047 http://www.wolframalpha.com/input/?i=%28111%2F570+-+.25%29%2Fsqrt%28.25%281-.25%29%2F570%29

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